(b) Looking at the horizontal forces at that point leads to 21sin35(49N)sin3528N.TT=°=°=(c) We denote the components of T3as Tx(rightward) and Ty(upward). Analyzing horizontal forces where string 2 and string 3 meet, we find Tx= T2= 28 N. From the vertical forces there, we conclude Ty= wB=50 N. Therefore,
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